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(A) For a compound microscope,the length of the tube $L$ for a relaxed eye (final image at infinity) is given by $L = v_o + f_e$,where $v_o$ is the im

Vedclass · Accepted Answer

$1.8$

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(A) For a compound microscope,the length of the tube $L$ for a relaxed eye (final image at infinity) is given by $L = v_o + f_e$,where $v_o$ is the image distance of the objective lens and $f_e$ is the focal length of the eyepiece.Given $L = 14 \, cm$ and $f_e = 5 \, cm$,we have $14 = v_o + 5$,which gives $v_o = 9 \, cm$.The magnifying power $m$ for a relaxed eye is given by $m = \frac{v_o}{u_o} 	imes \frac{D}{f_e}$,where $D$ is the least distance of distinct vision $(25 \, cm)$.Substituting the values: $25 = \frac{9}{u_o} 	imes \frac{25}{5}$.$25 = \frac{9}{u_o} 	imes 5$.$25 = \frac{45}{u_o}$.$u_o = \frac{45}{25} = 1.8 \, cm$.

The length of the compound microscope is $14 \, cm$. The magnifying power for a relaxed eye is $25$. If the focal length of the eye lens is $5 \, cm$,then the object distance for the objective lens will be.......$ cm$.

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